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Q.

A long container has air enclosed inside at room temperature and atmospheric pressure 105N/m2 between two movable pistons. It has a volume 20,000 cc. The area of cross section is 100  cm2 and force constant of springs is k=781.25  N/m2. Now push the right piston isothermally and slowly till it reaches the original position of the left piston-1 which is movable. What is the final length of air column. Assume that spring is initially relaxed and length of spring is very large as compare to separation between two piston.

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a

100 cm

b

50 cm

c

75 cm 

d

200 cm

answer is A.

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Detailed Solution

P1V1=P2V2 
(105)(20,000×106)={105+(2kcos2370)x100×104}(100×104)x 
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x2+x2=0 
x = 1 m = 100 cm

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